Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 33496 by malwaan last updated on 17/Apr/18

$$\mathrm{prove}\mathrm{that}$$$${\mathrm{e}}^{\mathrm{i}\pi }+1=0$$

Answered by MJS last updated on 17/Apr/18

$$\mathrm{just}\mathrm{use}\mathrm{the}\mathrm{definition}\mathrm{of}z\in \mathbb{C}:$$$$z=r{\mathrm{e}}^{\phi \mathrm{i}}=r\cos \left(\phi \right)+r\sin \left(\phi \right)\mathrm{i}\Rightarrow$$$$\Rightarrow {\mathrm{e}}^{\pi \mathrm{i}}=1\cos \left(\pi \right)+1\sin \left(\pi \right)\mathrm{i}=$$$$=-1+0\mathrm{i}=-1$$$$-1+1=0$$

Commented by malwaan last updated on 18/Apr/18

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com